If $f$ is continuous, $\Omega$ is open, $I=\{t:f(t)\in\Omega\}$ and $\sigma=\inf I^c$, does it hold $f(\sigma)\in\partial\Omega$?

Question

Let $\tau>0$, $E$ be a $\mathbb R$-Banach space, $\Omega\subseteq E$ be open, $f:[0,\tau]\to E$ be continuous with $f(0)\in\Omega$, $$I:=\{t\in[0,\tau]:f(t)\in\Omega\}$$ and $$\sigma:=\inf([0,\tau]\setminus I).$$

Assume $I\ne[0,\tau]$. Can we show that $\sigma\not\in I$ and $f(\sigma)\in\partial\Omega$?

Since $I\ne[0,\tau]$, we know that $\sigma\in[0,\tau]$ and hence $f(\sigma)$ is well-defined. Moreover, since $\Omega$ is open and $f$ is continuous, $I$ is $[0,\tau]$-open.

So, I've tried to assume the contrary, i.e. $\sigma\in I$, so that there would be a $\varepsilon>0$ with $B_\varepsilon(\sigma)\cap[0,\tau]\subseteq I$. This would yield that $[0,\tau]\setminus I\subseteq[0,\tau]\setminus B_\varepsilon(\sigma)$, but I wasn't able to derive a contradiction to the definition of $\sigma$ from that.

Answer 1

By definition $$[0, \tau] \setminus I = \lbrace t \in [0, \tau] : f(t) \notin \Omega \rbrace = f^{-1}(E \setminus \Omega)$$

so $$\sigma = \inf \left( f^{-1}(E \setminus \Omega)\right)$$

Because $\Omega$ is open in $E$, then $E \setminus \Omega$ is closed. By continuity, $f^{-1}(E \setminus \Omega)$ is closed in $[0, \tau]$, hence it is a compact. So $\sigma$ is the infimum of a compact, so it is a minimum, which means that $\sigma \in f^{-1}(E \setminus \Omega)$, so $\sigma \notin I$.

Now, for all $n$ sufficiently large, $\sigma - 1/n$ belongs to $[0, \tau]$ (because $f(0) \in \Omega$, so $\sigma > 0$), and by definition of $\sigma$ as the infimum, one has $f(\sigma - 1/n) \in \Omega$. But by continuity, $f(\sigma) = \lim f(\sigma - 1/n)$, so $f(\sigma)$ belongs to the closure of $\Omega$. Because $\Omega$ is open, and $f(\sigma) \notin \Omega$, you have indeed $f(\sigma) \in \partial \Omega$.